関数 $f(x_1, x_2, x_3) = e^{x_1} \sin x_2 \cos x_3$ に対して、点 $c = (\frac{\pi}{3}, \frac{\pi}{4}, 0)$ におけるテイラー展開を考えます。$h = (h_1, h_2, h_3)$ とします。以下の2つの量を計算します。 (1) $\sum_{\alpha \in \mathbb{Z}^3_+, |\alpha| = 1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$ (2) $\sum_{\alpha \in \mathbb{Z}^3_+, |\alpha| = 2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$

1. 問題の内容

関数 $f(x_1, x_2, x_3) = e^{x_1} \sin x_2 \cos x_3$ に対して、点 $c = (\frac{\pi}{3}, \frac{\pi}{4}, 0)$ におけるテイラー展開を考えます。$h = (h_1, h_2, h_3)$ とします。以下の2つの量を計算します。

(1) $\sum_{\alpha \in \mathbb{Z}^3_+, |\alpha| = 1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$

(2) $\sum_{\alpha \in \mathbb{Z}^3_+, |\alpha| = 2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$

2. 解き方の手順

(1) $|\alpha| = 1$ のとき、$\alpha$ は $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ のいずれかです。したがって、

$\sum_{\alpha \in \mathbb{Z}^3_+, |\alpha| = 1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{\partial f}{\partial x_1}(c) h_1 + \frac{\partial f}{\partial x_2}(c) h_2 + \frac{\partial f}{\partial x_3}(c) h_3$

まず偏微分を計算します。

$\frac{\partial f}{\partial x_1} = e^{x_1} \sin x_2 \cos x_3$

$\frac{\partial f}{\partial x_2} = e^{x_1} \cos x_2 \cos x_3$

$\frac{\partial f}{\partial x_3} = -e^{x_1} \sin x_2 \sin x_3$

次に $c = (\frac{\pi}{3}, \frac{\pi}{4}, 0)$ での値を計算します。

$\frac{\partial f}{\partial x_1}(c) = e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \cos 0 = e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} \cdot 1 = \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}}$

$\frac{\partial f}{\partial x_2}(c) = e^{\frac{\pi}{3}} \cos \frac{\pi}{4} \cos 0 = e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} \cdot 1 = \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}}$

$\frac{\partial f}{\partial x_3}(c) = -e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \sin 0 = -e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} \cdot 0 = 0$

したがって、

$\sum_{\alpha \in \mathbb{Z}^3_+, |\alpha| = 1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}} h_1 + \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}} h_2 + 0 \cdot h_3 = \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}} (h_1 + h_2)$

(2) $|\alpha| = 2$ のとき、$\alpha$ は $(2,0,0)$, $(0,2,0)$, $(0,0,2)$, $(1,1,0)$, $(1,0,1)$, $(0,1,1)$ のいずれかです。したがって、

$\sum_{\alpha \in \mathbb{Z}^3_+, |\alpha| = 2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{1}{2} \frac{\partial^2 f}{\partial x_1^2}(c) h_1^2 + \frac{1}{2} \frac{\partial^2 f}{\partial x_2^2}(c) h_2^2 + \frac{1}{2} \frac{\partial^2 f}{\partial x_3^2}(c) h_3^2 + \frac{\partial^2 f}{\partial x_1 \partial x_2}(c) h_1 h_2 + \frac{\partial^2 f}{\partial x_1 \partial x_3}(c) h_1 h_3 + \frac{\partial^2 f}{\partial x_2 \partial x_3}(c) h_2 h_3$

まず2階偏微分を計算します。

$\frac{\partial^2 f}{\partial x_1^2} = e^{x_1} \sin x_2 \cos x_3$

$\frac{\partial^2 f}{\partial x_2^2} = -e^{x_1} \sin x_2 \cos x_3$

$\frac{\partial^2 f}{\partial x_3^2} = -e^{x_1} \sin x_2 \cos x_3$

$\frac{\partial^2 f}{\partial x_1 \partial x_2} = e^{x_1} \cos x_2 \cos x_3$

$\frac{\partial^2 f}{\partial x_1 \partial x_3} = -e^{x_1} \sin x_2 \sin x_3$

$\frac{\partial^2 f}{\partial x_2 \partial x_3} = -e^{x_1} \cos x_2 \sin x_3$

次に $c = (\frac{\pi}{3}, \frac{\pi}{4}, 0)$ での値を計算します。

$\frac{\partial^2 f}{\partial x_1^2}(c) = e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \cos 0 = \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}}$

$\frac{\partial^2 f}{\partial x_2^2}(c) = -e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \cos 0 = -\frac{\sqrt{2}}{2} e^{\frac{\pi}{3}}$

$\frac{\partial^2 f}{\partial x_3^2}(c) = -e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \cos 0 = -\frac{\sqrt{2}}{2} e^{\frac{\pi}{3}}$

$\frac{\partial^2 f}{\partial x_1 \partial x_2}(c) = e^{\frac{\pi}{3}} \cos \frac{\pi}{4} \cos 0 = \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}}$

$\frac{\partial^2 f}{\partial x_1 \partial x_3}(c) = -e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \sin 0 = 0$

$\frac{\partial^2 f}{\partial x_2 \partial x_3}(c) = -e^{\frac{\pi}{3}} \cos \frac{\pi}{4} \sin 0 = 0$

したがって、

$\sum_{\alpha \in \mathbb{Z}^3_+, |\alpha| = 2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{1}{2} \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}} h_1^2 - \frac{1}{2} \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}} h_2^2 - \frac{1}{2} \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}} h_3^2 + \frac{\sqrt{2}}{2} e^{\frac{\pi}{3}} h_1 h_2 + 0 + 0 = \frac{\sqrt{2}}{4} e^{\frac{\pi}{3}} (h_1^2 - h_2^2 - h_3^2 + 2 h_1 h_2)$

3. 最終的な答え

(1) $\frac{\sqrt{2}}{2} e^{\frac{\pi}{3}} (h_1 + h_2)$

(2) $\frac{\sqrt{2}}{4} e^{\frac{\pi}{3}} (h_1^2 - h_2^2 - h_3^2 + 2 h_1 h_2)$